1) One of the valence electrons in a ground state atom has the quantum numbers n=4, l=1, ml=0, ms=-1/2. This atom could be:
a. Na
b.Cl
c. Rb
d. Br
e. Sn

2) What is the wavelength ofa beam of electrons travelling at 2.0*10^6 cm/s?
a. 6.33 cm
b.2.75*10^5 cm
c. 3.16*10^-36 cm
d. 3.64*10^-6 cm
2.43*10^-10 cm

3) Ca(s)+2HNO3 (aq) ----> Ca(NO3)2 (aq) + H2 (g)
A 0.500 g sample of Ca is treated with excess nitric acid and the hydrogen gas produced is bubbled through water into an inverted jar at a temperature of 60°C. The vapour pressure of water at 60°C is 149.4 mmHg. If the total pressure in the jar is 555 mmHg, what is the total volume of the gas?
a. 17.3 L
b. 8.52 L
c. 0.4567 L
d. 0.6939 L
e. 0.852 L

4) 6CO (g) + 13H2 (g) ----> C6H14 (g) + 6H2O (g)
what volume in L of the excess reactant remains if 29.7 L CO and 38.1 L H2 are allowed to react. Assume the volume of both gases are measured at 743°C and 1.29 atm.

1) If n= 4 ,l= 1
it means last shell is 4p
so atom should be of 4 th period
only Br atom is in fourth period and has 4p as the last valence orbital

2)
wavelength = h/mv (de-Broglie's relation )
h= 6.63 X 10^-34 Kg m^2 s^-1
m= 9.1 X 10^-31kg
v= 2.0 X 10 ^-6 cm/sec= 2.0 X 10^-8 m/sec
now just plug in all values in above equation.

3
first take out moles of H2 gas formed
moles of H2 = 0.500g Ca X 1mole of Ca/ 40.00g of Ca   X 1mole of H2 / 1mole of Ca = 0.0125 moles of H2
Now Pressure due to dry gas = (555-149.4)mm of Hg  = 405.6 mm of Hg X 1 atm / 760 mm of Hg
T = 60 + 273 = 333K

R= 0.0821L/mol.K
n= 0.0125 moles
PV=nRT
V= nRT/P
just plug in all values to get the answer

4)
Calculate  the number of moles of of H2 gas =PV/RT
Similarly calculate number of moles of CO = PV/RT
  Now find out the  excess reagent
Take out the left amount of excess reagent( number of moles ) and convert it into volume
       n1/V1 = n2 / V2
Rest we will discuss in the session